Researchers at a large nutrition and weight management company are trying to build a model to predict a person’s body fat percentage from an array of variables such as body weight, height, and body me

The correct answer is b. {eq}0.162 pm 0.525 {/eq}. The coefficient of weight is given as -0.162 in the Coefficients table. The standard error for this coefficient is also given as 1.355 in the same table. Using a 90% confidence level, we need to find the critical value for the t-distribution with 46 degrees of freedom (which is the degrees of freedom for the residual error). This value can be obtained from a t-table or a calculator and is \approx imately 1.677. Using this critical value and the standard error, we can calculate the margin of error as follows: margin of error = critical value * standard error = 1.677 * 1.355 = 2.29 Therefore, the 90% confidence interval for the coefficient of weight is: {eq}-0.162 pm 2.29 {/eq} Simplifying this expression, we get: {eq}-0.162 pm 0.525 {/eq} So, the correct answer is b. {eq}0.162 pm 0.525 {/eq}.